# 18.4 Electric field: concept of a field revisited  (Page 2/5)

 Page 2 / 5
$E=|\frac{F}{q}|=k|\frac{\text{qQ}}{{\mathrm{qr}}^{2}}|=k\frac{|Q|}{{r}^{2}}.$

Since the test charge cancels, we see that

$E=k\frac{|Q|}{{r}^{2}}.$

The electric field is thus seen to depend only on the charge $Q$ and the distance $r$ ; it is completely independent of the test charge $q$ .

## Calculating the electric field of a point charge

Calculate the strength and direction of the electric field $E$ due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge.

Strategy

We can find the electric field created by a point charge by using the equation $E=\text{kQ}/{r}^{2}$ .

Solution

Here $Q=2\text{.}\text{00}×{\text{10}}^{-9}$ C and $r=5\text{.}\text{00}×{\text{10}}^{-3}$ m. Entering those values into the above equation gives

$\begin{array}{lll}E& =& k\frac{Q}{{r}^{2}}\\ & =& \left(\text{8.99}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/C}}^{2}\right)×\frac{\left(\text{2.00}×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{C}\right)}{\left(\text{5.00}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{m}{\right)}^{2}}\\ & =& \text{7.19}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C.}\end{array}$

Discussion

This electric field strength is the same at any point 5.00 mm away from the charge $Q$ that creates the field. It is positive, meaning that it has a direction pointing away from the charge $Q$ .

## Calculating the force exerted on a point charge by an electric field

What force does the electric field found in the previous example exert on a point charge of $–0.250\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ ?

Strategy

Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field $\mathbf{\text{E}}=\mathbf{\text{F}}/q$ rearranged to $\mathbf{\text{F}}=q\mathbf{\text{E}}$ .

Solution

The magnitude of the force on a charge $q=-0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{μC}$ exerted by a field of strength $E=7\text{.}\text{20}×{\text{10}}^{5}$ N/C is thus,

$\begin{array}{lll}F& =& -\text{qE}\\ & =& \left(\text{0.250}×{\text{10}}^{\text{–6}}\phantom{\rule{0.25em}{0ex}}\text{C}\right)\left(7.20×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}\right)\\ & =& \text{0.180 N.}\end{array}$

Because $q$ is negative, the force is directed opposite to the direction of the field.

Discussion

The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.

## Phet explorations: electric field of dreams

Play ball! Add charges to the Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.

## Section summary

• The electrostatic force field surrounding a charged object extends out into space in all directions.
• The electrostatic force exerted by a point charge on a test charge at a distance $r$ depends on the charge of both charges, as well as the distance between the two.
• The electric field $\mathbf{\text{E}}$ is defined to be
$\mathbf{\text{E}}=\frac{\mathbf{\text{F}}}{q,}$

where $\mathbf{\text{F}}$ is the Coulomb or electrostatic force exerted on a small positive test charge $q$ . $\mathbf{\text{E}}$ has units of N/C.

• The magnitude of the electric field $\mathbf{\text{E}}$ created by a point charge $Q$ is
$\mathbf{\text{E}}=k\frac{|Q|}{{r}^{2}}.$

where $r$ is the distance from $Q$ . The electric field $\mathbf{\text{E}}$ is a vector and fields due to multiple charges add like vectors.

## Conceptual questions

Why must the test charge $q$ in the definition of the electric field be vanishingly small?

Are the direction and magnitude of the Coulomb force unique at a given point in space? What about the electric field?

## Problem exercises

What is the magnitude and direction of an electric field that exerts a $2\text{.}\text{00}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{N}$ upward force on a $–1.75\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge?

What is the magnitude and direction of the force exerted on a $3.50\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge by a 250 N/C electric field that points due east?

$8\text{.}\text{75}×{\text{10}}^{-4}$ N

Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff).

(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m?

(a) $6\text{.}\text{94}×{\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}\text{C}$

(b) $6\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}\text{N/C}$

Calculate the initial (from rest) acceleration of a proton in a $5\text{.}\text{00}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{N/C}$ electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.

(a) Find the direction and magnitude of an electric field that exerts a $4\text{.}\text{80}×{\text{10}}^{-\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}$ westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?

(a) $\text{300}\phantom{\rule{0.25em}{0ex}}\text{N/C}\phantom{\rule{0.25em}{0ex}}\left(\text{east}\right)$

(b) $4\text{.}\text{80}×{\text{10}}^{-\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}\phantom{\rule{0.25em}{0ex}}\left(\text{east}\right)$

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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